Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:

Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.

Let the five multiples be $7,14,21,28$ and 35 .

The cubes of these numbers are:

$7^{3}=343,14^{3}=2744,21^{3}=9261,28^{3}=21952$, and $35^{3}=42875$

Now, write the above cubes as a multiple of 73. Proceed as follows:

$343=7^{3} \times 1$

$2744=14^{3}=14 \times 14 \times 14=(7 \times 2) \times(7 \times 2) \times(7 \times 2)=(7 \times 7 \times 7) \times(2 \times 2 \times 2)=7^{3} \times 2^{3}$

$9261=21^{3}=21 \times 21 \times 21=(7 \times 3) \times(7 \times 3) \times(7 \times 3)=(7 \times 7 \times 7) \times(3 \times 3 \times 3)=7^{3} \times 3^{3}$

$21952=28^{3}=28 \times 28 \times 28=(7 \times 4) \times(7 \times 4) \times(7 \times 4)=(7 \times 7 \times 7) \times(4 \times 4 \times 4)=7^{3} \times 4^{3}$

$42875=35^{3}=35 \times 35 \times 35=(7 \times 5) \times(7 \times 5) \times(7 \times 5)=(7 \times 7 \times 7) \times(5 \times 5 \times 5)=7^{3} \times 5^{3}$

Hence, the cube of multiple of 7 is a multiple of 73.